# Probability of mean and sample variance.

Let $X_1,X_2$ and $X_3$ be a random sample from $N(3,12)$ distribution. If $\bar X=\dfrac{1}{3} \sum_{i=1}^{3}X_i$ and $S^2=\dfrac{1}{2}\sum_{i=1}^{3}(X_i-\bar X)^2$ denote the sample mean and the sample variance respectively, then $P(1.65<\bar X \leq4.35 ,.12<S^2 \leq55.26)$ is

$(A)=.49$

$(B)=.50$

$(C)=.98$

$(D)={}$None of the above

My first question is what does **

$P(1.65<\bar X \leq4.35 ,.12< S^2\leq55.26)$

** mean? Is it $P(1.65<\bar X \leq4.35 ,.12<S^2 \leq55.26)=P(1.65<\bar X \leq4.35)\cap P(.12<S^2 \leq55.26)$?

I have doubt in this very first step also i have done some work after that assuming(the step above mentioned) and i want to ask about that too but that i will ask if this step is fine. I will edit my question but before that please tell me about it. (My English is weak so please ignore grammar mistakes).

As the sample are independent identically normally distributed r.v.s, thus $S^2 = \frac{1}{n-1}\sum (X_i - \bar{X})^2$ ans $\bar{X}_n$ are independent as well (this is not a trivial claim, but I guess in your context you can use it as a fact) and recall that $$\bar{X}_n \sim \mathcal{N}(\mu, \sigma^2/n), \quad \frac{S^2}{\sigma^2} \sim \frac{1}{n-1}\chi^2_{(n-1)}.$$ And recall that if $A$ and $B$ are independent events then $$\mathbb{P}(A\cap B)= \mathbb{P}(A)\mathbb{P}(B).$$
• Well, if the r.v.s are independent then $P(1.65<\bar X \leq4.35 ,12<S^2 \leq55.26)=P(1.65<\bar X \leq4.35) P(12<S^2 \leq55.26)$ – V. Vancak Jan 29 '18 at 12:28
• Thank you very much i had doubt in that portion. I am more familiar with $\cap$ between two variables. – Daman Jan 29 '18 at 12:50